Extremizing the boundary value problem $$I[y]=\int_0^1y'^2(x)\,dx+y^2(0)-2y^2(1)$$
My Thought:
First, we use Euler-Lagrange equation and solving we get , $y(x)=C_1x+C_2$. Then we put it in $I[y]$ which gives a function of $C_1$ and $C_2$. Now for extreme value we put $\frac{\partial I}{\partial C_1}=0$ and $\frac{\partial I}{\partial C_2}=0$. These two relations give $C_1=C_2=0$. Then we get $y(x)=0$, which is trivial..
Is my process right ? I think it is wrong..
If you calculate $I[y]$ for $y(x)=C_1x+C_2$, you will find that it equals $$ -C_1^2-4C_1C_2-C_2^2. $$ In particular, if $C_1=C_2$, you get $I[y]=-6C_1^2$, so the functional is not bounded below. On the other hand, if $C_1=-C_2$, then $I[y]=2C_1^2$, so the functional is not bounded above either.